Question: Simplify the following expression and state the condition under which the simplification is valid: $z = \dfrac{a^2 + 6a - 7}{a^2 + 4a - 21}$
First factor the expressions in the numerator and denominator. $ \dfrac{a^2 + 6a - 7}{a^2 + 4a - 21} = \dfrac{(a - 1)(a + 7)}{(a - 3)(a + 7)} $ Notice that the term $(a + 7)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(a + 7)$ gives: $z = \dfrac{a - 1}{a - 3}$ Since we divided by $(a + 7)$, $a \neq -7$. $z = \dfrac{a - 1}{a - 3}; \space a \neq -7$